How Do You Calculate Energy and Speed on a Roller Coaster?

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Homework Help Overview

The discussion revolves around calculating energy and speed on a roller coaster, specifically focusing on kinetic and potential energy at different points of a loop. The original poster presents a scenario involving a person traveling at a specific speed and height, prompting calculations related to energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of kinetic and potential energy formulas, questioning the correct use of variables and arithmetic. There is an attempt to clarify the relationship between energy at the top and bottom of the loop, with some participants suggesting the conservation of energy principle.

Discussion Status

The discussion is active, with participants providing feedback on calculations and encouraging clarification of concepts. Some guidance has been offered regarding the correct application of energy equations, and there is an ongoing exploration of how to determine speed at the bottom of the loop.

Contextual Notes

Participants are grappling with potential confusion regarding the role of friction and the correct setup of energy equations. There is a mix of correct and incorrect calculations being shared, highlighting the need for careful consideration of units and variables.

alijan kk
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Homework Statement


Ima Scaarred (m=56.2 kg) is traveling at a speed of 12.8 m/s at the top of a 19.5-m high roller coaster loop.

a. Determine Ima's kinetic energy at the top of the loop.
b. Determine Ima's potential energy at the top of the loop.
c. Assuming negligible losses of energy due to friction and air resistance, determine Ima's total mechanical energy at the bottom of the loop (h=0 m).
d. Determine Ima's speed at the bottom of the loop.[/B]

Homework Equations


PE=MGH
KE=0.5*MV^2[/B]

The Attempt at a Solution


i tried to find b

pe=0.5*56.2*12.8

=14,027.52help me understanding the question.

the answers are.

  • Show Answer
a. 4.60 x 103 J
b. 1.07 x 104 J
c. 1.53 x 104 J
d. 23.4 m/s
 
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alijan kk said:
i tried to find b
pe=0.5*56.2*12.8

=14,027.52
Looks like you've mixed your two relevant equations together.

your relevant equation says pe = mgh, but that's not what you've done, what is m? g? h?
First identify all your variables so you know where they'll go in your equations.

Also, your arithmetic is wrong, 0.5*56.2*12.8 does not equal 14,027.52
Show your units with your working, and be specific on exactly where you're stuck/ what you don't understand, it makes it easier to help.
 
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billy_joule said:
Looks like you've mixed your two relevant equations together.

your relevant equation says pe = mgh, but that's not what you've done, what is m? g? h?
First identify all your variables so you know where they'll go in your equations.

Also, your arithmetic is wrong, 0.5*56.2*12.8 does not equal 14,027.52
Show your units with your working, and be specific on exactly where you're stuck/ what you don't understand, it makes it easier to help.
PE= mass*gravity*height

i want to imagine the picture of that question so that it would be easy for me , i want to understand the question
 
sorry the answer of pe=0.5*56.2*12.8
is 359.68
but it is also wrong...

is there any role of friction?
or use of this formula
Pi+Ki=Pf+Pi
 
All the ANswers i have got! thank you bilijoule i was making big mistake,,,,
 
alijan kk said:
All the ANswers i have got! thank you bilijoule i was making big mistake,,,,

Great, good work.
 
billy_joule said:
Great, good work.
thankyou\
but i m still trying to find the speed at the bottom of the loop .

KE=KE
4600=0.5*56.2*v^2
4600=28.1*v^2
4600/28.1=v^2
v^2=163.7
square root both sides
v=12.8

so this way i got the initial velocity but how to get the speed at the bottom?
 
When Ima gets to the bottom, the potential energy she had at the top has been converted to kinetic energy via Conservation of Energy.
This should be somewhat intuitive, ever ridden your bike down a hill? You don't even need to pedal to gain kinetic energy..it's coming from somewhere else..
The higher the hill the faster you'll be going at the bottom...
The relevant equation:
EKi + EPi = EKf + EPf

(which is what I think your earlier equation in post #4 was supposed to be?
Pi+Ki=Pf+Pi
)

Solving for EKf is the answer to c), then solving EKf = 1/2 mv2 for v will give the answer to d)
 
v=23 m/s ! thanks for the equation .
KEf=1/2mv^
 

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